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A box contains m white and n black balls. Balls are drawn at random one at a time without replacement. Find the probability of encountering a white ball by the kth draw.

This is our tutorial sheet's question and I don't know the answer. Thanks in advance.

Edit : I corrected the title and I misunderstood the question. I read 'at' instead of 'by'. With 'by', the question is very easy, still thank you for your answers. If you have solution for 'at', please answer. Again thank you for your efforts.

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    (k-1Ca)/2 divided by total number of ways sum over a from 0 to k, if k is less than m, i dont know any easy simple form – 2017-02-04
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    If there is a lottery where $m$ of the $m+n$ lots carries a price. If you take part in this then what is your chance on a price? You might have been the $k$-th person buying a lot, but does that really matter? – 2017-02-04

4 Answers 4

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The problem is to find the probability of getting a white ball in $k$-th draw. May be for the first time or may be not... so we may have already gotten a white ball in first $k-1$ times. Therefore we have to consider them. If we consider '$i$' as no. of times a black ball can come in those $k-1$ times(as $k$-th is assumed to be a white ball) $i=0,1,2,.....k-1$.

Total probability is : $(m/m+n) + (n/m+n)(m/m+n-1) + (n/m+n)(n-1/m+n-1)(m/m+n-2) + (n/m+n)(n-1/m+n-1)(n-2/m+n-2)(m/m+n-3) + ...+ (n/m+n)(n-1/m+n-1)(n-2/m+n-2)...(n-k+2/m+n-k+2)(m/m+n-k+1)$

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Hint: what is the probability of only having drawn black balls the first $k$ draws?

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    Your title asks about white ball on the $k$th draw, while the body of your question asks about white ball _by_ the $k$th draw. This hint is for the latter. – 2017-02-04
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    Thank you for your answer. Can you please read the Edit I added? – 2017-02-04
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    The probability of drawing a ball at the $k$th draw is actually the same as for the first draw. You can just accept that there is nothing inherently different about the different draws (as long as we don't know anything about the other draws), or you can start the really long calculation of conditioned probabilities and work it out that way. It's up to you. – 2017-02-04
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The probability of encountering a white ball by the $k^{th}$ is equal to $1 - P$ where $P$ is the probability that the first $k$ balls were all black. The latter probability is easier to compute. For example, the probability of the first ball being black is $\frac{n}{m+n}$. The probability that the second ball is also black is $\frac{n-1}{m+n-1}$. So, the probability that the first two balls are both black is $\frac{n(n-1)}{(m+n)(m+n-1)}$. The probability that the first $k$ balls are all black is similarly given by $\frac{n(n-1)...(n-k+1)}{(m+n)(m+n-1)...(m+n-k+1)}$.

Note that if $k > n$, then the probability becomes 1 because clearly you will have to draw a white ball by the $k^{th}$ draw (a version of Pigeon Hole Principle).

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    Thank you for your answer. Can you please read the Edit I added? – 2017-02-04
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    The probability of picking a white ball at the $k^{th}$ try is again $\frac{m}{m+n}$. It does not really matter which try because imagine laying down the balls in a row from left to right. Whichever position you pick, the probability that you have a white ball in that spot is the same and is m/m+n. Hope that clarifies things. – 2017-02-05
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Hint (if you are searching for the probability that the $k$-th draw results in a white ball).

Suppose that all ball are taken out and one of them (randomly taken) is handed over to you. What is the probability that you will receive a white ball?

In the same line:

Suppose that for $i=1,2,\dots,m+n$ we have a person $i$.

The $m+n$ balls are taken out of the box and are randomly handed over to the persons.

Each of them receives a ball.

What is the probability that the person with number $k$ will receive a white ball?

Think of the answer on the first question posed in this answer and indentify with person $k$.

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    Thank you for your answer. Can you please read the Edit I added? – 2017-02-04